Sphere Equation Analysis

The question asks for the diameter of a sphere given its equation: \(2x^2+2y^2 + 2z^2 + 3x + 3y+3z-6=0\). To find the diameter, we first need to determine the sphere's radius using the standard equation of a sphere.

Standard Sphere Equation

The standard equation of a sphere with center \((a, b, c)\) and radius \(r\) is:

The diameter (\(D\)) of the sphere is twice the radius, so \(D = 2r\).

Converting the Given Sphere Equation

The given equation is \(2x^2+2y^2 + 2z^2 + 3x + 3y+3z-6=0\). We need to rearrange this into the standard form.

1. Divide by the coefficient of squared terms: Since the coefficients of \(x^2\), \(y^2\), and \(z^2\) are all 2, we divide the entire equation by 2 to make these coefficients 1: \(x^2 + y^2 + z^2 + \frac{3}{2}x + \frac{3}{2}y + \frac{3}{2}z - 3 = 0\)
2. Complete the square: We group the \(x\), \(y\), and \(z\) terms and complete the square for each. The general form for completing the square for \(t^2 + bt\) is \((t + \frac{b}{2})^2 - (\frac{b}{2})^2\).
   • For \(x\): \(x^2 + \frac{3}{2}x = (x + \frac{3}{4})^2 - (\frac{3}{4})^2 = (x + \frac{3}{4})^2 - \frac{9}{16}\)
   • For \(y\): \(y^2 + \frac{3}{2}y = (y + \frac{3}{4})^2 - (\frac{3}{4})^2 = (y + \frac{3}{4})^2 - \frac{9}{16}\)
   • For \(z\): \(z^2 + \frac{3}{2}z = (z + \frac{3}{4})^2 - (\frac{3}{4})^2 = (z + \frac{3}{4})^2 - \frac{9}{16}\)
3. Substitute back and rearrange: Substitute these completed square forms back into the equation: \((x + \frac{3}{4})^2 - \frac{9}{16} + (y + \frac{3}{4})^2 - \frac{9}{16} + (z + \frac{3}{4})^2 - \frac{9}{16} - 3 = 0\) Now, move the constant terms to the right side of the equation: \((x + \frac{3}{4})^2 + (y + \frac{3}{4})^2 + (z + \frac{3}{4})^2 = 3 + \frac{9}{16} + \frac{9}{16} + \frac{9}{16}\)
4. Simplify the right side: Combine the constant terms on the right side: \(3 + \frac{27}{16} = \frac{3 \times 16}{16} + \frac{27}{16} = \frac{48}{16} + \frac{27}{16} = \frac{75}{16}\)

The equation of the sphere in standard form is:

Calculating Sphere Radius and Diameter

By comparing this standard form to \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\), we can identify:

• The center of the sphere is \((a, b, c) = (-\frac{3}{4}, -\frac{3}{4}, -\frac{3}{4})\).
• The square of the radius is \(r^2 = \frac{75}{16}\).

Now, we find the radius \(r\):

Finally, we calculate the diameter (\(D\)) of the sphere:

Therefore, the diameter of the sphere is \(\frac{5\sqrt{3}}{2}\).